Two small conducting spheres of equal radius have charges `+ 10 muC` and `-20 muC` respectively and placed at a distance R from each other. They experience force `F_(1)`. If they are brought in contact and separated to the same distance, they experience force `F_(2)`. The ratio of `F_(1)` to `F_(2)` is

Options

A. `1:2`
B. `-8:1`
C. `8:1`
D. `-2:1`

Answer

B

Explanation

Here, `F_(1)=(k(+10)(-20))/(R^(2))=(-kxx200)/(R^(2))" ….(i)"`
As spheres have of equal radii and their capacities are same.
`therefore` On touching, the net charge `=(10-20)muC=-10muC,` is sphared equally between them, i.e. each spheres carries `-5muC` charge.
`F_(2)=(k(-5)(-5))/(R^(2))`
`F_(2)=(kxx 25)/(R^(2))" ...(ii)"`
Divide Eq. (i) by (ii), we get
`(F_(1))/(F_(2))=(-8)/(1) rArr F_(1):F_(2)=-8:1`